A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be

Engineering

Physics

Center of Mass

Work Done by a Constant Force

Forces and Types of Forces

Question

A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its ${(\frac{1}{n})}^{th}$ part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be

$\frac{2 MgL}{n^{2}}$

nMgL

$\frac{MgL}{2 n^{2}}$

$\frac{MgL}{n^{2}}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$Mass of the hanging part = \frac{M}{n}$

$h_{COM} = \frac{L}{2 n}$

Work done $W = {mgh}_{COM} = (\frac{M}{n}) g (\frac{L}{2 n}) = \frac{MgL}{2 n^{2}}$